7–84. Evaluate the following integrals.
77. ∫ arccosx dx

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Identify the integral to solve: \(\int \arccos x \, dx\).

Use integration by parts, which states: \(\int u \, dv = uv - \int v \, du\). Choose \(u = \arccos x\) and \(dv = dx\).

Compute \(du\) by differentiating \(u\): since \(u = \arccos x\), then \(du = -\frac{1}{\sqrt{1 - x^2}} \, dx\). Also, integrate \(dv\) to get \(v = x\).

Apply the integration by parts formula: \(\int \arccos x \, dx = x \arccos x - \int x \left(-\frac{1}{\sqrt{1 - x^2}}\right) dx = x \arccos x + \int \frac{x}{\sqrt{1 - x^2}} \, dx\).

Evaluate the remaining integral \(\int \frac{x}{\sqrt{1 - x^2}} \, dx\) by using a substitution such as \(w = 1 - x^2\), then express the integral in terms of \(w\) and solve.

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주요 개념

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Integration by Parts

Integration by parts is a technique used to integrate products of functions. It is based on the product rule for differentiation and follows the formula ∫u dv = uv - ∫v du. Choosing appropriate u and dv simplifies the integral, especially when dealing with inverse trigonometric functions.

Derivative of the Inverse Cosine Function

The derivative of arccos(x) is -1 / √(1 - x²). Knowing this derivative is essential when applying integration by parts, as it helps determine du when u = arccos(x). This derivative reflects the rate of change of the inverse cosine function.

Basic Integration Techniques

Understanding fundamental integration methods, such as integrating powers and roots, is important for solving the resulting integrals after applying integration by parts. This includes recognizing standard integral forms and manipulating expressions to fit these forms.

7–84. Evaluate the following integrals.77. ∫ arccosx dx